Is there a necessary and sufficient condition for a linearly ordered topological space to be pseudo-metrizable? (by a pseudo-metric, I mean a map $X\times X\rightarrow\mathbb{R}$ in which all the metric axioms are satisfied except that the distance between two distinct points may be zero)

A negative answer is, of course, still acceptable; I was wondering if there is an analogue of Urysohn's metrization theorem with respect to pseudo-metrics in linearly ordered spaces.

nota metric can never be Hausdorff, so there can be no linearly ordered space which is strictly pseudo-metric. $\endgroup$ – Simon_Peterson Apr 2 '16 at 8:24